2e^2.1

that's two e to the power of 2.1

does this equal 16.33 (2dp) ?

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- Aug 28th 2012, 03:07 AM #1

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- Aug 28th 2012, 03:13 AM #2
## Re: 2e^2.1

Hint: $\displaystyle \displaystyle \begin{align*} e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \dots \end{align*}$

Substitute $\displaystyle \displaystyle \begin{align*} x = 2.1 \end{align*}$ and evaluate to as much accuracy as you desire.

- Aug 28th 2012, 03:59 AM #3

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- Aug 28th 2012, 04:03 AM #4

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- Aug 28th 2012, 10:06 AM #5

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- Aug 28th 2012, 09:30 PM #6

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## Re: 2e^2.1

Here's another perfectly fine way of doing this. You know how to take integer powers and suppose you have $\displaystyle e$ handy, to good precision. What can we do then? We have

$\displaystyle 2e^{2.1} = 2e^2 e^{1/10}$

Now you can use the Taylor series for that last exponential

$\displaystyle e^{1/h} = 1+\frac{1}{h}+\frac{1}{2h^2}+\frac{1}{6}h^3 + \cdots$

If you have $\displaystyle 2e^2$ to several decimal places in precision, then you can get away with using less terms in this series. It's not a big save but it's kind of neat.

- Aug 29th 2012, 02:54 AM #7

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- Aug 31st 2012, 05:55 AM #8

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- Aug 31st 2012, 06:02 AM #9

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